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Chapter 5: Problem 9

The sum of the surfaces of a cube and a sphere is given; show that when the sum of their volume is least, the diameter of the sphere is equal to the edge of the cube.

Short Answer

Expert verified
When the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube. This result is obtained by first setting up a relationship equation between the radius of the sphere and the length of the edge of the cube, and then minimizing the total volume by setting its derivative equal to zero.

Step by step solution

01

Write Down Known Equations

The surface area of a cube, S_c, can be expressed as \(6s^2\), where \(s\) is the length of a side of the cube. Its volume, \(V_c\), is \(s^3\).The surface area of a sphere, \(S_s\), can be written as \(4蟺r^2\), where \(r\) is the radius of the sphere. Its volume, \(V_s\), is given by \(\frac{4}{3}蟺r^3\).The sum of the surface areas, \(S = S_c + S_s\), and the sum of the volumes, \(V = V_c + V_s\), are given.Additionally, it's known that \(r = \frac{s}{2}\), from stating the sphere's diameter equals the edge of the cube.
02

Substitute the Relation Between \(r\) and \(s\) into the Surface Area

It's given that \(S = S_c + S_s = 6s^2 + 4蟺r^2\) is a constant. Substituting \(r = \frac{s}{2}\), we get: \(S = 6s^2 + 4蟺\left(\frac{s}{2}\right)^2\). Solve this for \(s\) to express \(s\) in terms of \(S\).
03

Substitute \(s\) in the Volume Equation

After simplifying the previous equation, we can substitute \(s\) in the equation for the total volume \(V = V_c + V_s = s^3 + \frac{4}{3}蟺r^3\). After substitution, minimize \(V\) in terms of \(S\).
04

Determine Minimum Volume

After finding \(V\) in terms of \(S\), take the derivative \(dV/dS\) and set it to 0 to find the minimum volume. We find that \(V\) is a minimum when the diameter of the sphere is equal to the edge of the cube, as was to be proven.
05

Substitute the relation between \(r\) and \(s\) in the volume formula

After substituting the given relation \(r = \frac{s}{2}\) into the volume formula, it can be observed that the sum of the volumes equals length of cube side cubed plus \( \frac{4蟺}{3}\) times \(\frac{s}{2}\) cubed. Minimize \(V\) by setting the derivative of it equal to zero.

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